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Serway and Vuille - College Physics 12/e (Homework)

James Finch

Physics - Labs, section 1, Fall 2019

Instructor: Sarah Anders

Current Score : 19 / 64

Due : Saturday, February 17, 2035 11:00 EST

Last Saved : n/a Saving...  ()

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1. 1/1 points  |  Previous Answers SerCP12 4.STEP.4.1A. My Notes
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Points
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1
1/1
6/100
Total
1/1
 
  • This is an example of a STEP module. STEP is a scaffolded series of questions, organized into levels, modules and topics, which helps students achieve multi-concept mastery, build confidence, and work at their own pace. With each question, a "STEP back" button is offered which opens an alternate version of a question from a prerequisite skill. It is from Topic 4: Newton's Laws of Motion and Module 4: Tension Forces, Levels 1–5.

A vertical string supports a block having a mass of 5.00 kg as shown in the figure.
A string hangs down from a horizontal surface. A block of mass m is attached to the lower end of the string.
Calculate the magnitude of the tension force (in N) exerted by the string on the block. (The local free-fall acceleration is 9.80 m/s2.)
Correct: Your answer is correct. seenKey

49

 N


Solution or Explanation
The free-body diagram for the block is shown below.
A free-body diagram shows two vectors. An x y-coordinate plane indicates that the +x-axis is to the right and the +y-axis is up along the page.
  • Vector T points upward.
  • Vector Fg = m vector g points downward.
The lengths of vector T and vector Fg are approximately equal.
Only two forces act on the block: the tension force T exerted by the string and the force of gravity,
Fg = −mg.
 Because the acceleration in the y-direction is zero,
ay = 0,
 and the y-component of the second law yields
Fy = may
T − mg = 0
T = mg = (5.00 kg)(9.80 m/s2)
T = 49.0 N.
Important Concepts/Remarks
The tension force is due to electromagnetic forces associated with atoms and molecules in the string.
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2. 1/1 points  |  Previous Answers SerCP12 4.STEP.4.2A. My Notes
Question Part
Points
Submissions Used
1
1/1
3/100
Total
1/1
 
  • This is an example of a STEP module. STEP is a scaffolded series of questions, organized into levels, modules and topics, which helps students achieve multi-concept mastery, build confidence, and work at their own pace. With each question, a "STEP back" button is offered which opens an alternate version of a question from a prerequisite skill. It is from Topic 4: Newton's Laws of Motion and Module 4: Tension Forces, Levels 1–5.

A block having a mass of 40.0 kg and attached by a wire to a winch accelerates straight upward as shown in the figure.
A wire hangs down from a winch. A block of mass m is attached to the lower end of the wire. Vector a points upward.
If the acceleration of the block is 1.50 m/s2, find the magnitude of the tension in the wire (in N). (The local free-fall acceleration is 9.80 m/s2.)
N
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3. 1/1 points  |  Previous Answers SerCP12 4.STEP.4.3A. My Notes
Question Part
Points
Submissions Used
1
1/1
3/100
Total
1/1
 
  • This is an example of a STEP module. STEP is a scaffolded series of questions, organized into levels, modules and topics, which helps students achieve multi-concept mastery, build confidence, and work at their own pace. With each question, a "STEP back" button is offered which opens an alternate version of a question from a prerequisite skill. It is from Topic 4: Newton's Laws of Motion and Module 4: Tension Forces, Levels 1–5.

A crate having a mass of 80.0 kg is supported by two identical ropes, each making an angle of 64.0° with respect to the vertical as shown in the figure.
Two ropes hang down from a horizontal surface. A crate of mass m is attached to the lower ends of the ropes. The ropes go up and left and up and right, respectively, from the crate to the horizontal surface, each making acute angle θ with the vertical.
Calculate the magnitude of the tension force in each of the ropes (in N). The local free-fall acceleration is 9.80 m/s2.
Correct: Your answer is correct. seenKey

894

 N


Solution or Explanation
The free-body diagram for the crate is shown below.
A free-body diagram shows three vectors. An x y-coordinate plane indicates that the +x-axis is to the right and the +y-axis is up along the page.
  • Vector T points up and left at acute angle θ from the vertical.
  • A second vector T points up and right at angle θ from the vertical.
  • Vector Fg = m vector g points downward.
Vector Fg is longer than each individual vector T.
Three forces act on the crate: the tension forces exerted by the ropes and the force of gravity. By symmetry, the tension forces in the ropes are identical. The part of each tension force that acts in the positive y-direction is
T cos(θ).
 Because
ay = 0,
 the y-component of the second law yields
Fy = may
2T cos(θ) − mg = 0.
Rearrange this expression algebraically for the tension T.
T = 
mg
2 cos(θ)
 = 
(80.0 kg)(9.80 m/s2)
2 cos(64.0°)
T = 894 N
Important Concepts/Remarks
Whether the sine or cosine is used to give a vector component in a given direction depends on the angle supplied in the problem. Here, the adjacent side of the triangle formed by the vectors and the coordinate directions is parallel to the y-axis, so the cosine is required. Any cables, strings, or wires attached to objects in such problems must be symmetrically arranged, or else the tensions will not be equal.
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4. 2/2 points  |  Previous Answers SerCP12 4.STEP.4.4A. My Notes
Question Part
Points
Submissions Used
1 2
1/1 1/1
4/100 2/100
Total
2/2
 
  • This is an example of a STEP module. STEP is a scaffolded series of questions, organized into levels, modules and topics, which helps students achieve multi-concept mastery, build confidence, and work at their own pace. With each question, a "STEP back" button is offered which opens an alternate version of a question from a prerequisite skill. It is from Topic 4: Newton's Laws of Motion and Module 4: Tension Forces, Levels 1–5.

A wrecking ball having a mass of 750 kg is supported by a tension force T1 supplied by a horizontal cable in the negative x-direction and the tension T2 of a second cable that makes an angle of 30.0° with respect to the vertical (or 60.0° with respect to the positive x-axis; see the figure).
A wrecking ball has mass m. An x y-coordinate plane indicates that the +x-axis is to the right and the +y-axis is up along the page. Two forces act on the wrecking ball.
  • Force T1 pulls the wrecking ball to the left.
  • Force T2 pulls the wrecking ball up and right, at acute angle θ to the vertical.
Calculate the magnitude of the tension forces T1 and T2 (in N). The local free-fall acceleration is 9.80 m/s2. (Treat the object as a point particle.)
T1= Correct: Your answer is correct. seenKey

4240

 N T2= Correct: Your answer is correct. seenKey

8490

 N


Solution or Explanation
Note: We are displaying rounded intermediate values for practical purposes. However, the calculations are made using the unrounded values.
The free-body diagram for the wrecking ball is shown below.
A free-body diagram shows three vectors. An x y-coordinate plane indicates that the +x-axis is to the right and the +y-axis is up along the page.
  • Vector T1 points to the left.
  • Vector T2 points up and right, making acute angle θ with the vertical.
  • Vector Fg = m vector g points downward.
Vector T2 is the longest of the three vectors.
Three forces act on the wrecking ball: the tension force T1 exerted by the horizontal cable, the tension force T2 by the cable at the given angle, and gravity,
Fg = −mg.
Step 1: Write the y-component of Newton's Second Law
Because the given angle (30.0°) is with respect to the vertical, the adjacent side of the triangle formed by the vector T with arrow2 corresponds to the y-direction, requiring a cosine function. Because
ay = 0,
 the y-component of the second law yields
Fy = may
−mg + T2 cos(θ) = 0
Step 2: Solve for the Tension T2
T2 = 
mg
cos(θ)
 = 
(750 kg)(9.80 m/s2)
cos(30.0°)
T2 = 8,490 N
Step 3: Write the x-component of Newton's Second Law and Solve for T1
Because the given angle 30.0° is with respect to the vertical, the opposite side of the triangle formed by the vector T2 corresponds to the x-direction, requiring a sine function. Because the wrecking ball is not accelerating in the x-direction,
ax = 0.
 The x-component of the second law can therefore be solved for the tension T1.
Fx = max
−T1 + T2 sin(θ) = 0
T1 = T2 sin(θ) = (8,490 N) sin(30.0°)
T1 = 4,240 N
Important Concepts/Remarks
In this case a single tension, T2, supports the object against gravity, making for an easy solution of two equations in two unknowns.
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5. 2/2 points  |  Previous Answers SerCP12 4.STEP.4.5A. My Notes
Question Part
Points
Submissions Used
1 2
1/1 1/1
1/100 1/100
Total
2/2
 
  • This is an example of a STEP module. STEP is a scaffolded series of questions, organized into levels, modules and topics, which helps students achieve multi-concept mastery, build confidence, and work at their own pace. With each question, a "STEP back" button is offered which opens an alternate version of a question from a prerequisite skill. It is from Topic 4: Newton's Laws of Motion and Module 4: Tension Forces, Levels 1–5.

A block having a mass of 14.0 kg is supported by a tension force T1 supplied by a string making an angle
ϕ = 40.0°
 with respect to the vertical and the tension T2 supplied by a second string making an angle θ = 30.0° with respect to the vertical. (See the figure; not to scale.)
A block has mass m. Two forces act on the block.
  • Force T1 pulls the block up and left, at acute angle ϕ to the vertical.
  • Force T2 pulls the block up and right, at acute angle θ to the vertical.
Calculate the magnitude of the tension forces T1 and T2 (in N). The local free-fall acceleration is 9.80 m/s2.
T1= Correct: Your answer is correct. seenKey

73

 N T2= Correct: Your answer is correct. seenKey

93.9

 N


Solution or Explanation
Note: We are displaying rounded intermediate values for practical purposes. However, the calculations are made using the unrounded values.
The free-body diagram for the block is shown below.
A free-body diagram shows three vectors. An x y-coordinate plane indicates that the +x-axis is to the right and the +y-axis is up along the page.
  • Vector T1 points up and left, making acute angle ϕ with the vertical.
  • Vector T2 points up and right, making acute angle θ with the vertical.
  • Vector Fg = m vector g points downward.
Three forces act on the block: the tension force T1 exerted by the string at the angle
ϕ = 40.0°,
 the tension force T2 exerted by the string at the angle
θ = 30.0°,
 and gravity,
Fg = −mg.
Step 1: Write the x-component of the Second Law
Because the angles are given with respect to the vertical, the opposite sides of the triangles formed by each vector and the coordinate axes correspond to the x-components of those vectors, requiring sine functions. Because
ax = 0,
 the x-component of the second law yields
Fx = max
−T1 sin(ϕ) + T2 sin(θ) = 0.
Step 2: Rearrange Algebraically and Solve for T1
T1 = 
T2 sin(θ)
sin(ϕ)
Because both angles are known, it is somewhat simpler, algebraically, to substitute for them.
T1 = 
sin(30.0°)
sin(40.0°)
T2
T1 = 0.778T2
            (1)
Step 3: Write the x-component of the Second Law
Because the angles are given with respect to the vertical, the adjacent sides of the triangles formed by each vector and the coordinate axes correspond to the y-components of those vectors, requiring cosine functions. Because
ay = 0,
 the y-component of the second law yields
Fy = may
−mg + T1 cos(ϕ) + T2 cos(θ) = 0.
             (2)
Step 4: Substitute the Expression for T1 and Solve for T2
Substitute the expression for T1 from equation (1) into equation (2).
−mg + 0.778T2 cos(ϕ) + T2 cos(θ) = 0
Add mg to both sides and factor T2.
0.778 cos(ϕ) + cos(θ)
T2 = mg
T2 = 
mg
0.778 cos(ϕ) + cos(θ)
 = 
(14.0 kg)(9.80 m/s2)
0.778 cos(40.0°) + cos(30.0°)
T2 = 93.9 N
Step 5: Substitute the Quantity Found for T2 into Equation (1)
T1 = 0.778T2 = (0.778)(93.9 N)
T1 = 73.0 N
Important Concepts/Remarks
This problem requires solving two linear equations in two unknowns. Although the substitution method was used, it is also possible to use Gaussian elimination or Cramer's Rule, for example.
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6. 1/2 points  |  Previous Answers SerCP12 5.6.P.023.MI. My Notes
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1 2
0/1 1/1
1/100 1/100
Total
1/2
 
  • Master It tutorials are an optional student-help tool available within select questions for just-in-time support. Students can use the tutorial to guide them through the problem-solving process step-by-step using different numbers.

A 1,900 kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.00 m before coming into contact with the top of the beam, and it drives the beam 13.2 cm farther into the ground as it comes to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.
magnitude Incorrect: Your answer is incorrect. seenKey

5.83e+05
Write an expression for the work done by all of the forces acting on the beam as it moves into the ground and relate that work to the change in energy of the pile driver in order to find the force that the beam exerts on the pile driver. N direction Correct: Your answer is correct. seenKey

upward

Need Help? Master It

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7. 9/9 points  |  Previous Answers SerCP12 13.A.P.069.MI.SA. My Notes
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1/1 1/1 1/1 1/1 1/1 1/1 1/1 1/1 1/1
2/100 1/100 1/100 1/100 1/100 1/100 1/100 1/100 2/100
Total
9/9
 
  • Master It tutorials—Standalone are embedded, step-by-step tutorials used to help students understand each step required to solve the problem, before inputting their final answer.

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part.

Tutorial Exercise
A large block P executes horizontal simple harmonic motion as it slides across a frictionless surface with a frequency
f = 1.45 Hz.
 Block B rests on it, and the coefficient of static friction between the two is
μs = 0.545.
 What maximum amplitude of oscillation can the system have if block B is not to slip?
Two blocks are stacked, with a smaller block B on top of a larger block P, which is on top of a horizontal surface. The coefficient of static friction between B and P is
μs.
 The left side of block P is attached to the right end of a horizontal spring, and the left end of the spring is attached to a wall.
Step 1
Let's consider a system that consists only of block B. We'll first determine the forces acting on this system. Let's assume that the spring is compressed, so that the point where the spring is attached to block P is to the left of the relaxed position. Let's also assume for the moment that block B does not slip on block P as it moves.
Which of the following forces act on block B? Select all that apply.
Correct: Your answer is correct.

✓ Correct. The only forces acting on block B are contact forces from block P - the normal and static friction forces - and its weight. The static friction force is to the right, because block P is to the left of the equilibrium position. Just as the spring force to the right will slow down block P and reverse its direction, so does the static friction force to block B.
Step 2
The free body diagram below shows the forces acting on block B when the spring is compressed. n with arrow is the normal force, f with arrows is the static friction force, and mg with arrow is the weight of block B (m is its mass).
A free-body diagram shows three forces acting on block B, here shaped like a cube.
  • Force vector n acts upward on the top face.
  • Force vector fs acts to the right on the bottom face.
  • Force vector Fg = m vector g acts downward on the block.
To find the maximum amplitude before block B starts to slip, we need to consider the maximum static friction force. Recall that the static friction force can vary up to a maximum value. As block B oscillates back and forth, the static friction force will oscillate as well. At which point (or points) in the block's motion will the static friction force be at its largest magnitude?
     Correct: Your answer is correct.
✓ Correct. In simple harmonic motion (sinusoidal motion), the restoring force - in this case, the static friction force - is at the maximum displacements from the equilibrium point. (That is, at the points corresponding to positive or negative amplitude.)
When the static friction force is at its largest value, which of the following quantities is also at a maximum?
     Correct: Your answer is correct.
✓ Correct. In simple harmonic motion (sinusoidal motion), the maximum restoring force - in this case, the static friction force - is at the maximum displacements from the equilibrium point. (That is, at the points corresponding to positive or negative amplitude.)
Step 3
If the acceleration at the maximum displacement points is too high, the static frictional force on block B will exceed the maximum value, and the contact force between the two blocks will shift from static to kinetic. The block will then begin to slip. So, we need to determine the maximum static friction force and relate it to the maximum acceleration, which in turn we will relate to the amplitude (the maximum displacement).
So, let's first consider the magnitude of the maximum static friction force.
From Newton's second law applied to the vertical forces in the free body diagram, write an expression for the magnitude of the normal force on block B. Your answer should be in terms of mass m and g.
n =
$$mg
Correct: Your answer is correct. webMathematica generated answer key Correct. If +y is upward, then Newton's second law gives
Fy
 = n − mg = 0,
 since the acceleration is only in the x direction. Therefore
n = mg.
Step 4
From the definition of the static friction force and the above result, write an expression for the maximum static friction force magnitude in terms of the coefficient of static friction
μs,
 the mass m, and g.
fs,max =
$$μs​mg
Correct: Your answer is correct. webMathematica generated answer key Correct. The static friction force is related to the normal force. If +y is upward, then Newton's second law gives
fs ≤ μsn,
 so the maximum static friction force is
fs,max = μsn
 Since we found
n = mg
 above,
fs,max = μsmg.
Step 5
Applying Newton's second law in the horizontal direction, write a symbolic expression for the maximum acceleration of block B. Your answer should only be in terms of
μs,
 m, and g (but you might not use all of these symbols).
amax =
$$μs​g
Correct: Your answer is correct. webMathematica generated answer key Correct. From Newton's second law,
Fx
 = fs,max − mamax.
Solving for
amax
gives
amax = 
fs
m
.
 Substituting in the result from Step 4 above, we find the mass m divides out, leaving
amax = μsg.
Step 6
Now let's relate the acceleration to the amplitude by considering the harmonic oscillation (that is, sinusoidal motion) of block B. If block B oscillates without slipping, which of the following expressions relates its acceleration at any time t to its amplitude A and frequency f? (Assume
t = 0
 when the block is at maximum displacement from equilibrium.)
     Correct: Your answer is correct.
✓ Correct. If the block is at maximum displacement at
t = 0,
 then displacement is
x = Acos(2πft),
 velocity is
v = A(tπf)sin(2πft)
 and acceleration is
a = −A(2πf)2cos(2πft).
We're interested in the maximum magnitude of the acceleration. From the above result, which of the following relates the maximum acceleration of block B to its amplitude A and frequency f?
     Correct: Your answer is correct.
✓ Correct. Because
−1 < cos(2πft) < 1,
 the maximum magnitude of the acceleration corresponds to
|cos(2πft)| = 1.
 From the correct expression above, we find
amax = A(2πf)2.
Step 7
We found an expression for the maximum acceleration in terms of static friction in step 5 and an expression for it in terms amplitude in step 6. We can now equate these expressions and solve for the maximum amplitude. Write an expression for the maximum amplitude in terms of
μs,
 g, and f.
A =
$$gμs​(2πƒ)2​
Correct: Your answer is correct. webMathematica generated answer key Correct. This is the maximum possible amplitude before slipping, given the coefficient of static friction
μs
 and the frequency of oscillation
f.
Using the given values of
f = 1.45 Hz
 and
μs = 0.545,
 you should verify that this expression gives a maximum amplitude of
A = 0.0643 m,
 or 6.43 cm.
Conclusion
We applied Newton's second law and the equation for the maximum static frictional force to find an expression for the maximum acceleration of the top block before slipping between the two blocks occurs. We then substituted this expression into the equation for acceleration of a sinusoidal oscillation in terms of its amplitude and angular frequency. We used the given frequency of oscillation and the coefficient of static friction between the two blocks to find the maximum amplitude of oscillation that the two-block system can attain before slippage occurs between the blocks.
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8. 1/9 points  |  Previous Answers SerCP12 5.DWB.006. My Notes
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1 2 3 4 5 6 7 8 9
–/1 –/1 1/1 –/1 –/1 –/1 –/1 –/1 –/1
0/100 0/100 2/100 0/100 0/100 0/100 0/100 0/100 0/100
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1/9
 
  • Digital Workbook questions interweave narrative and interactives of a variety of media types with assessments, containing an introduction to the relevant topics and subtopics.

Conservation of Energy
Now that we've introduced both kinetic and potential energy, we're ready to discuss in more detail the principle of conservation of energythe law of physics which states that energy cannot be created or destroyed, but only change forms; more precisely, the total energy of the universe remains constant.. Watch the video below, then answer the questions that follow it.
Part 1 of 5
(a)
Which of the following is true about a closed system?
    
(b)
Consider a system consisting of a falling ball and the Earth. While the ball is falling, what is happening to the potential energy, kinetic energy, and total energy of this system?
    
Part 2 of 5
The principle of conservation of energy says that energy cannot be created out of nothing, nor can it simply disappear. When we are dealing with a closed systemany system that does not interact with its environment, which is also sometimes called an "isolated system", the total energy of that system must remain the same. Remember that a closed system is one where there is no interaction with the environment (no net force, or no transfer of matter or energy).
Expressed mathematically, we write
Ef = Ei
 (for a closed system),
or the final energy of the system equals its initial energy. We can also consider the change in the energy of a closed system, in which case the equation is
Ef − Ei = ΔE = 0.
The change in energy ΔE is zero for a closed system.
However, as we saw in the video, energy in a closed system can change forms. The total energy is then really the sum of many different types of energy. Each of these energy types could presumably change, but the total stays the same.
What types of energy are included in the total? There can be many different types, some of which we've already studied, such as the following.
Kinetic Energy: the energy of motion of objects. To be more precise, we typically mean translational kinetic energy—the energy of the overall "straight-line" motion of an object. A rotating object also has rotational kinetic energy. (We won't go into details on how to calculate rotational kinetic energy.)
Potential energy: the energy of interaction between objects in the system, such as gravitational potential energy. Note that there can be potential energy associated with other types of interactions, such as electric potential energy.
Internal energy: This is really just a general category for any energy that is stored, microscopically, within the objects of the system. It could be further broken into the following.
  • Thermal energy: this is the energy of the random, microscopic motion of atoms and molecules making up objects. It is associated with temperature.
  • Chemical energy: the energy stored in the chemical bonds between atoms and molecules, which can change during chemical reactions.
  • Nuclear energy: the energy stored within the nucleus of atoms, associated with the forces that bind the protons and neutrons of nuclei together.
This seems like a lot to take into account. But in practice, when applying conservation of energy, we only need to worry about the types of energy that are changing during a process.
So, let's come back to our falling ball example, shown in the video. Which types of energy changed significantly while the ball was falling? Take the initial state of the system to be just after the ball is released, and the final state to be just before the ball hits the ground. (Select all that apply.)
Correct: Your answer is correct.

✓ Correct. Clearly the height above the ground changes, so the gravitation potential energy changes. The ball speeds up, so the kinetic energy changes. The ball does not undergo any chemical or nuclear reactions, so those types of energy do not change. And while it is true that air resistance does increase the ball's temperature slightly, the effect is negligible in this case, so we can approximate the thermal energy change as zero. Note, however, in many cases thermal energy change due to air resistance or friction cannot be ignored. We'll discuss this in a later section.
Part 3 of 5
Let's now work through an example using conservation of energy.
Let's say a small, rubber ball with a mass of 1.00 kg is launched straight up in the air from ground level by a toy compressed-air cannon. Assume that the cannon is small, so we'll consider the initial height of the ball to be zero when it leaves the cannon. The initial speed of the ball when it leaves the cannon is 11.5 m/s.
Our goal is to find the maximum height the ball reaches before it begins to fall.
You may have solved a similar type of problem when studying kinematics. But now, let's apply conservation of energy to this problem.
If we assume, reasonably, that air resistance is minimal, then our ball–Earth system is a closed system. That means the total energy of the system is constant, or
Ef = Ei,
the final total energy is equal to the initial total energy.
Let's think about what "final" and "initial" refer to in this case. Where is the ball in the "initial" situation in this problem?
    
The only types of energy that are changing in this situation are the kinetic energy of the ball and the gravitational potential energy. We can then expand the total energy, on each side of the equation, into two parts—the kinetic energy
EK
 and the potential energy
EP.
 The equation then becomes
EK,f + EP,f = EK,i + EP,i.
In other words, the sum of the final kinetic and potential energies is equal to the sum of the initial kinetic and potential energies.
Which of the terms in the above equation do we know are equal to zero? (Select all that apply.)



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–/5
 
  • Interactive Video Vignettes target common student misconceptions with video-analysis and interactive modules that focus on learning difficulties.

Watch the video vignette below, and be sure to follow any instructions when prompted during the videos. Then answer the questions below about the concepts discussed in the video.
(a)
As you saw in the video, the two bullets in the experiment were fired from the same rifle. Assume the bullets also had the same mass and were fired from identically prepared cartridges. From this, what can we conclude about the initial momentum of the bullets just before they struck the blocks?
    
(b)
As you saw in the video, the two blocks had about the same mass, and both were initially at rest. Which of the following is true about the two bullet–block systems, just after the bullets collided with their respective blocks?
    
(c)
Which of the following statements provides correct reasoning for why both bullet–block systems reached the same maximum height?
    
(d)
Which of the following statements are true about the energy of the two bullet–block systems immediately after the collisions? (Select all that apply.)

(e)
Imagine comparing the two blocks after the experiment. In which block would we expect the bullet to be embedded the greatest distance?
    
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  • Optimized Problems are designed to combat academic honesty issues by providing numerical and contextual variation to randomize the problem with comprehensive written solutions and targeted feedback.

In a scene from a television show, a car rolls down an incline and off a vertical cliff, falling into a valley below. The car starts from rest and rolls down the incline, which makes an angle of 24.0° below the horizontal, with a constant acceleration of 3.82 m/s2. After rolling down the incline a distance of 35.0 m, it reaches the edge of the cliff, which is 40.0 m above ground level.
(a)
How much time (in s) does it take the car to fall from the edge of the cliff to the landing point?
s
(b)
At the point where the car crashes into the ground, how far is it horizontally from the edge of the cliff (in m)?
m
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  • Read It links are available as a learning tool under each question so students can quickly jump to the corresponding section of the eTextbook.

A uniform plank of length 2.00 m and mass 32.5 kg is supported by three ropes, as indicated by the blue vectors in the figure below. Find the tension in each rope when a 690–N person is d = 0.500 m from the left end.
A person stands on a horizontal 2.00 m long plank a distance d from the left end. The plank is supported by three ropes.
  • A rope at the right end of the plank goes up and right, making a 40.0° angle with the horizontal and applying a force vector T1 on the plank outward along the rope.
  • A rope at the left end of the plank goes vertically up, applying a force vector T2 on the plank outward along the rope.
  • A rope at the left end of the plank goes horizontally left, applying a force vector T3 on the plank outward along the rope.
magnitude of
T1
     		N
magnitude of
T2
     		N
magnitude of
T3
     		N
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  • Active Examples build a bridge between practice and homework by guiding students through the process needed to master a concept and include worked-out solutions.

EXAMPLE 8.2 The Swinging Door
Goal Apply the more general definition of torque.

(a) Top view of a door being pushed by a 300-N force. (b) The components of the 300-N force.
Figures a and b illustrate the top view of a door such that the hinge is on the left and the doorknob is on the right. The door has a width of 2.00 m and the point O is the position of the hinge.
  • In figure a, on the right side of the door, a vector of magnitude 300 N points up and to the right making an angle of 60.0° with the horizontal.
  • In figure b, on the right side of the door, an arrow labeled 260 N points up and an arrow labeled 150 N points to the right.
Problem   (a)   A man applies a force of F = 3.00 102 N at an angle of 60.0° to the door of Figure (a), 2.00 m from well-oiled hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation.   (b)   Suppose a wedge is placed 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won't open the door?

Strategy Part (a) can be solved by substitution into the general torque equation. In part (b) the hinges, the wedge, and the applied force all exert torques on the door. The door doesn't open, so the sum of these torques must be zero, a condition that can be used to find the wedge force.
SOLUTION
(a) Compute the torque due to the applied force exerted at 60.0°.
Substitute into the general torque equation.
τF = rFsin θ = (2.00 m)(3.00 ✕ 102 N) sin 60.0°
 = (2.00 m)(2.60 ✕ 102 N) = 5.20 102 N · m
(b) Calculate the force exerted by the wedge on the other side of the door.
Set the sum of the torques equal to zero.
τhinge + τwedge + τF = 0
The hinge force provides no torque because it acts at the axis (r = 0). The wedge force acts at an angle of −90.0°, opposite the upward 260 N component.
0 + Fwedge(1.50 m) sin (−90.0°) + 5.20 ✕ 102 N · m = 0
Fwedge = 347 N
LEARN MORE
Remarks Notice that the angle from the position vector to the wedge force is −90°. This is because, starting at the position vector, it's necessary to go 90° clockwise (the negative angular direction) to get to the force vector. Measuring the angle in this way automatically supplies the correct sign for the torque term and is consistent with the right-hand rule. Alternately, the magnitude of the torque can be found and the correct sign chosen based on physical intuition. Figure (b) illustrates the fact that the component of the force perpendicular to the lever arm causes the torque.

Question To make the wedge more effective in keeping the door closed, should it be placed closer to the hinge or to the doorknob?
    

PRACTICE IT
Use the worked example above to help you solve this problem.
(a) A man applies a force of F = 3.00 102 N at an angle of 60.0° to a door, x = 2.30 m from the hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation.
N · m

(b) Suppose a wedge is placed 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won't open the door?
N
EXERCISE HINTS:  GETTING STARTED  |  I'M STUCK!
A man ties one end of a strong rope 8.95 m long to the bumper of his truck, 0.522 m from the ground, and the other end to a vertical tree trunk at a height of 3.70 m. He uses the truck to create a tension of 8.94 102 N in the rope. Compute the magnitude of the torque on the tree due to the tension in the rope, with the base of the tree acting as the reference point.
N · m
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  • Pre-Lecture Explorations visually prepare your students for your next lecture with laddered modules that enable students to interact with the topics and reinforce conceptual and analytical understanding.

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part.

Prelecture Exploration: Energy Conservation for Rolling Objects
Darcel and Chandra are sitting in the park after physics class, and they notice some children rolling various objects down a slight incline in the sidewalk. Darcel is curious how the masses and shapes of the objects affect their motion, and Chandra says that it might be fun to think about the energy of motion associated with rolling objects. The friends decide to use the simulation to explore the motion and energies of various objects rolling down an incline.

They can click on an object (solid sphere, spherical shell, hoop, or cylinder) to position it at the top of the incline and click "roll" to release it. The translational and rotational speeds of the object and the time interval the object is on the incline are shown, together with a graph of the percentages of the different energy types as the object rolls down the incline. The "roll" button turns into a "pause" button while the object is in motion to allow them to examine different parameters at various points. For the object–Earth system in the simulation, the zero configuration of gravitational potential energy is defined to occur when the object is at the bottom of the incline. For the simulation, air resistance and any rolling friction effects are neglected, and the masses of all the objects are equal.

Click here to open the simulation in a new window.
Part 1 of 11 - Comparison of Similar Objects of Different Sizes
Darcel clicks on the large solid sphere, then clicks "roll." He pays attention to the time the sphere takes to roll down the incline. Then he clicks on the small solid sphere and repeats. Which of Darcell's statements about his observations of the time it takes for the two objects to reach the bottom is correct?
    


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  • Tutorial questions include multi-step tutorial used to help students understand each step required to solve the problem.

Tutorial 8.2: Applying the Rotational Second Law The rotational analog of Newton's second law is
τ
 = Iα,
where all torques
τ,
 moments of inertia
I,
 and angular accelerations α, are computed around a fixed axis. This equation states that the angular acceleration of an extended rigid object is proportional to the net torque acting on it. The moment of inertia
I
 is the rotational analog of translational mass.

For a collection of point particles, the moment of inertia is given by
I = 
mr2.
 For rigid objects of uniform composition, the textbook lists moments of inertia about different axes.

Recall that the magnitude of torque generated by a force of magnitude
F
 is
τ = rF sin(θ),
where, as before,
r
 is the length of the position vector,
F
 is the magnitude of the applied force, and θ is the angle between the vectors
r
 and
F.
 By convention, when an applied force causes an object to rotate counterclockwise, the torque on the object is positive; when the force causes the object to rotate clockwise, the torque on the object is negative.

The Guided Problem guide you through the following Problem-Solving Strategy.
•   Read the problem carefully at least once.
•   Draw a picture of the system, identify the object of interest, and indicate forces with arrows.
•   Label each force in the picture.
•   Draw a free-body diagram of the object of interest.
•   Apply the rotational analog of Newton's second law.
•   Solve for the desired unknown quantity and substitute the numbers.
Guided Problem
A small sphere of mass
m = 2.20 kg
 is attached to the end of a very light, rigid rod of length
L = 1.30 m
 as shown in the figure. The mass and weight of the rod are negligible, and the sphere is small enough to be treated as a point particle. The left end of the rod is attached to the top of the pivot but free to rotate without friction around an axis perpendicular to the plane of the figure, and the sphere is held so the rod is horizontal and at rest. At the instant the sphere is released, determine its angular acceleration about an axis perpendicular to the plane of the figure and passing through the pivot point.
A horizontal rod of length L has a small sphere labeled m attached to its right end. Its left end rests on a pivot.
Part 1 of 6
Read the problem carefully at least once.
Be sure to notice which quantities are known and which must be found. The known quantities are the sphere's mass and the length of the rod. The rod's mass and weight are negligible.

The unknown quantity to be determined is the angular acceleration of the sphere at the instant it is dropped.

Identify all of the forces acting on the rod and the sphere shown in the figure.
    


Additional Problems

Question 8.2a:
The figure shows a block of mass
m = 2.60 kg
 hanging from a thin rope wrapped around a solid, disk-shaped pulley. The pulley has mass
M = 8.50 kg,
 radius
R = 0.129 m,
 and is free to rotate on a frictionless, horizontal axle. If the block is released from rest, determine the magnitude of its downward acceleration.
A pulley of radius R and mass M is shown. An object of mass m is attached to the end of a string and hangs over the right side of the pulley.
m/s2

Question 8.2b:
A simple yo-yo is made by wrapping a thin string of negligible mass around the perimeter of a solid disk of radius
R = 0.120 m
 and mass
M = 2.56 kg.
 The free end of the string is held stationary by a person's hand, as in the figure, and the yo-yo is released from rest. Determine the magnitude of the disk's downward acceleration.
A circle represents a yo-yo of radius R and mass M. A string wraps clockwise around the yo-yo, then extends straight upward from the leftmost point of the yo-yo. The other end of the string is held by a hand.
m/s2
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  • Conceptual Questions help diagnose student misconceptions and give them practice on qualitative reasoning.

A solid disk and a hoop are simultaneously released from rest at the top of an incline and roll down without slipping. Which object reaches the bottom first?
     Correct: Your answer is correct.
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